Question: A few families took a trip to an amusement park together. Tickets cost $$6.50$ each for adults and $$3.00$ each for kids, and the group paid $$53.00$ in total. There were $5$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Explanation: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${6.5x+3y = 53}$ ${x = y-5}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-5}$ for $x$ in the first equation. ${6.5}{(y-5)}{+ 3y = 53}$ Simplify and solve for $y$ $ 6.5y-32.5 + 3y = 53 $ $ 9.5y-32.5 = 53 $ $ 9.5y = 85.5 $ $ y = \dfrac{85.5}{9.5} $ ${y = 9}$ Now that you know ${y = 9}$ , plug it back into ${x = y-5}$ to find $x$ ${x = }{(9)}{ - 5}$ ${x = 4}$ You can also plug ${y = 9}$ into ${6.5x+3y = 53}$ and get the same answer for $x$ ${6.5x + 3}{(9)}{= 53}$ ${x = 4}$ There were $4$ adults and $9$ kids.